Kenneth Kuttler

13.5. THE DUAL SPACE OF C0 (X) 333

Then hi is continuous and

h1(x)+h2(x) = g(x), |hi| ≤ fi.

The function hi is clearly continuous at points x where f1 (x)+ f2 (x)> 0. The reason it iscontinuous at a point where f1 (x)+ f2 (x) = 0 is that at every point y where f1 (y)+ f2 (y)>0, the top description of hi gives

|hi (y)|=∣∣∣∣ fi (y)g(y)

f1 (y)+ f2 (y)

∣∣∣∣≤ |g(y)| ≤ f1 (y)+ f2 (y)

so if |y− x| is small enough, |hi (y)| is also small. Then it follows from this definition ofthe hi that

−ε +λ ( f1 + f2) ≤ |Lg|= |Lh1 +Lh2| ≤ |Lh1|+ |Lh2|≤ λ ( f1)+λ ( f2).

Since ε > 0 is arbitrary, this shows that

λ ( f1 + f2)≤ λ ( f1)+λ ( f2)≤ λ ( f1 + f2)

The last assertion follows from the observation that if |g| ≤ f , then ∥g∥∞≤ ∥ f∥

∞so

λ ( f ) = sup{|Lg| : |g| ≤ f} ≤ sup∥g∥∞≤∥ f∥∞

∥L∥∥g∥∞≤ ∥L∥∥ f∥

∞■

Let Λ be the unique linear extension of Theorem 7.7.7 for which Λ f = λ ( f ) when f ≥0. It is just like defining the integral for functions when you understand it for nonnegativefunctions. As with integrals Λ( f )≤ Λ(| f |) = λ (| f |). Then from the above lemma,

|Λ f | ≤ λ (| f |)≤ ∥L∥∥ f∥∞

. (13.7)

Also, if f ≥ 0,Λ f = λ ( f ) ≥ 0. Therefore, Λ is a positive linear functional on C0(X).In particular, it is a positive linear functional on Cc (X). Thus there are now two linearcontinuous mappings L,Λ which are defined on C0 (X) with the norm ∥·∥

∞. The above

13.7 shows that in fact ∥Λ∥ ≤ ∥L∥. Also, from the definition of Λ

|Lg| ≤ λ (|g|) = Λ(|g|)≤ ∥Λ∥∥g∥∞

so in fact, ∥L∥ ≤ ∥Λ∥ showing that these two have the same operator norms, ∥L∥= ∥Λ∥.By Theorem 8.2.1 on Page 188, since Λ is a positive linear functional on Cc (X), there

exists a unique measure µ such that Λ f =∫

X f dµ for all f ∈ Cc(X). This measure isregular. In fact, it is actually a finite measure. First note that by density of Cc (X) in C0 (X)

∥Λ∥ = sup{Λ f : f ∈Cc (X) ,∥ f∥∞≤ 1}= sup{Λ f : 0≤ f ≤ 1, f ∈Cc (X)}

= sup{∫

f dµ : 0≤ f ≤ 1, f ∈Cc (X)

}= µ (X)

This is stated in the following lemma.

Lemma 13.5.4 Let L ∈ C0 (X)′ as above. Then letting µ be the Radon measure justdescribed, it follows µ is finite and µ (X) = ∥Λ∥= ∥L∥ .

13.5. THE DUAL SPACE OF Cy (X) 333Then h; is continuous andhy (x) +ho(x) = g(x), Ail < fi.The function h; is clearly continuous at points x where f; (x) + f2 (x) > 0. The reason it iscontinuous at a point where f| (x) + f2 (x) = 0 is that at every point y where fi (y) + fo (y) >0, the top description of h; givesfily)g() |hi(y)| = |= | Sle I SAO) +A(9) = | pO | < le <0) +00)so if |y—x| is small enough, |/;(y)| is also small. Then it follows from this definition ofthe h; that—e+A(fit fr) [Lg] = |Lhy + Lha| < |Lhy| + |Lho|A(fi) +A(f2).<<Since € > 0 is arbitrary, this shows thatAfi + fr) SA(fi)+A(fr) SAA + fa)The last assertion follows from the observation that if |g| < f, then ||g||,. < || f||.. soA(f) = sup{|Lg| : |g] Sf} S er Ell IIslleo S NZI NIF leoBlloo STII llooLet A be the unique linear extension of Theorem 7.7.7 for which Af = A (f) when f >0. It is just like defining the integral for functions when you understand it for nonnegativefunctions. As with integrals A(f) < A(|f|) =A (|f|). Then from the above lemma,IAF] SA CIAL) SEM Files - (13.7)Also, if f >0,Af =A(f) > 0. Therefore, A is a positive linear functional on Co(X).In particular, it is a positive linear functional on C,(X). Thus there are now two linearcontinuous mappings L,A which are defined on Co (X) with the norm ||-||,,. The above13.7 shows that in fact ||A]] < ||Z||. Also, from the definition of A[Lg] <A (Ig|) =A(Ig]) < IAT Isle.so in fact, ||L]|| < ||A|| showing that these two have the same operator norms, ||Z|| = ||A\.By Theorem 8.2.1 on Page 188, since A is a positive linear functional on C; (X), thereexists a unique measure pt such that Af = fy fdu for all f € C.(X). This measure isregular. In fact, it is actually a finite measure. First note that by density of C, (X) in Co (X)sup {Af : f € Co (X), [If ll. <1} =sup{Af:0< f<l,fec.(X)}sup{ [ fanso<r<i.sec(x)}=wx)|AllThis is stated in the following lemma.Lemma 13.5.4 Let L € Co(X)! as above. Then letting be the Radon measure justdescribed, it follows is finite and U(X) = ||A|| = ||L||.